Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $z = \dfrac{y + 6}{y - 2} \times \dfrac{-4y^2 - 32y + 80}{y^2 - 3y - 54} $
Answer: First factor out any common factors. $z = \dfrac{y + 6}{y - 2} \times \dfrac{-4(y^2 + 8y - 20)}{y^2 - 3y - 54} $ Then factor the quadratic expressions. $z = \dfrac {y + 6} {y - 2} \times \dfrac {-4(y - 2)(y + 10)} {(y + 6)(y - 9)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {(y + 6) \times -4(y - 2)(y + 10) } {(y - 2) \times (y + 6)(y - 9) } $ $z = \dfrac {-4(y - 2)(y + 10)(y + 6)} {(y + 6)(y - 9)(y - 2)} $ Notice that $(y + 6)$ and $(y - 2)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {-4(y - 2)(y + 10)\cancel{(y + 6)}} {\cancel{(y + 6)}(y - 9)(y - 2)} $ We are dividing by $y + 6$ , so $y + 6 \neq 0$ Therefore, $y \neq -6$ $z = \dfrac {-4\cancel{(y - 2)}(y + 10)\cancel{(y + 6)}} {\cancel{(y + 6)}(y - 9)\cancel{(y - 2)}} $ We are dividing by $y - 2$ , so $y - 2 \neq 0$ Therefore, $y \neq 2$ $z = \dfrac {-4(y + 10)} {y - 9} $ $ z = \dfrac{-4(y + 10)}{y - 9}; y \neq -6; y \neq 2 $